Wikipedia:Reference desk/Archives/Mathematics/2008 March 27

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 27

Trigonometry

This is a homework question which I've tried but always left clueless, so I was hoping that someone here could be kind and help me out. 'A pyramid is 146m high. Person A & Person B are looking at the top of it. Angle of elevation from top of pyramid to B is 12 degrees. If the distance between Person A & B is 25m and both A & B are 1.8m tall, what is the distance from B to centre of base of pyramid?' Thanks

Algebra problem

I do not understand these kinds of problems. 4c to the 2nd power+4c+1 Thankyou —Preceding unsigned comment added by 64.119.61.7 (talk • contribs) 11:10, 27 March 2008

What's the problem? Finding roots of that quadratic? There are various techniques you can use - I would use factorisation for that one. --Tango (talk) 01:19, 27 March 2008 (UTC)

Algebra problem 2

The problem is p to the 2nd power+14p+49=0. I understand that you use 7,7. Now do I just right 7,7 for my answer or do I do this-(p+7)(p+7)? Thanks —Preceding unsigned comment added by 64.119.61.7 (talk • contribs) 11:13, 27 March 2008

The solution to p^2 + 14p + 49 = 0 is p = -7 (you missed the minus sign, but this time I'll give it to you). You generally don't have to list double roots twice, unless asked to (or if the fact that it's a double root is important). If you're asked to factorise the expression p^2 + 14p + 49, then you write (p + 7)^2 or (p + 7)(p + 7). If you're asked to solve it, then you would probably factorise it first anyway, and then use that to find the solutions - (p + a)(p + b) = 0 means that either p + a = 0 or p + b = 0, and then solve like any linear equation. The form the answer takes depends on the question. Confusing Manifestation(Say hi!) 00:19, 27 March 2008 (UTC)

antiderivative help on ?

A.) It says to first find f ' and then f f ''(x)=x^3+7x^2+8x-4, f '(1)=5, f(1)=-9 I have already figured out that f '(x)=(x^4/4)+7(x^3/3)+8(x^2/2)-4x+2.4166... but i cant figure out what f (x)=________________

B.) also i am having trouble with a similar one. it says: Given f ''(x)=cos(x), f '(pi/2)=8, f(pi/2)=8, find: f '(x):_________(which i have already calculated the correct anwser to be: sinx +7 f (x):__________________(this is the one that i again dont understand how to get.

Can anyone please help...people have tried helping but everytime i do it i still get it wrong. pls help!!

00:54, 27 March 2008 (UTC)Bubbles16x (talk) 00:54, 27 March 2008 (UTC)

(I've added nowiki tags so your double primes don't get turned into italics tags!) To get from f' to f you do exactly the same thing you did to get from f'' to f'. --Tango (talk) 01:14, 27 March 2008 (UTC)

And for doing that, it helps if you realize that

(x⁴/4) + 7(x³/3) + 8(x²/2) − 4x + 29/12 = (1/4)x⁴ + (7/3)x³ + (8/2)x² − 4x + 29/12.

For problem B, here is a method you can use for this and similar problems.

Step 1: Make a reasonable guess for the form of the answer. Since f'(x) has both a trig function part (namely cos(x)) and a polynomial part (namely 7), we make room for both. It takes some practice and trial and error to get good guesses. In this case you can use

f(x) = a sin(x) + b cos(x) + cx + d,

in which form the variables a, b, c and d are still unknown. You have to determine their values.

Step 2. From that formula, compute f'(x) = ... This will again be something with sines, cosines and a polynomial part.

Step 3. Now equate what you computed in Step 2 with what you already have:

... = sin(x) + 7.

Step 4. What does that tell you about a, b, c and d? How can you choose their values so that the two sides of the equation become equal? You should be able to find at least some values.

Step 5. Substitute the known values you have found for the variables in the form we guessed for f(x).

Finally. If you got that far, one unknown variable will still remain. Use the fact that f(π/2) = 8 to find its value and finish this off.  --Lambiam 07:05, 27 March 2008 (UTC)

The number line

On the number line, the segment from 0 to 1 are divided into x and y equal parts, respectively. For example, if x=2 and y=3, the points 1/3, 1/2 and 2/3 are marked. Given x=X and y=Y, besides manually calculating all such points, is there a formula that can calculate the least distance between any two marks? Imagine Reason (talk) 01:53, 27 March 2008 (UTC)

In other words, given x, y natural numbers >1, you want to know the minimum value of |ax-by| for a, b natural numbers with a less than y, b less than x. Isn't this just the HCF of x and y? Algebraist 02:46, 27 March 2008 (UTC)

I think the problem is more to find the minimum of |a/x − b/y|. If x and y have a common factor, some "x marks" will coincide with some "y marks". For example, if x = 6 and y = 10, we have 3/6 = 5/10. In general, if x = pz and y = qz, p/x = q/y. That gives a distance of 0 with this formula. Otherwise, the least distance is 1/(xy).

If coinciding "x marks" and "y marks" are not considered to be separate – in other words, we just take the set of all marks – we get a different result for when x and y have a common factor. The case x = y = 2 must be excluded, for in that case there is only 1 mark in total. Otherwise, the least distance is 1/m, where m is the least common multiple of x and y.  --Lambiam 07:39, 27 March 2008 (UTC)

I think you are both saying the same thing, since HCF(x,y) / xy = 1 / LCM(x,y). Algebraist is working in units of 1/xy. Gandalf61 (talk) 10:06, 27 March 2008 (UTC)

Yeah, I just prefered to think in terms of integers only. Algebraist 15:23, 27 March 2008 (UTC)

Could you please help me visualize why it's 1/xy or the HCF? I can see now the case where they share a common factor, but otherwise I still don't understand it. Thank you. Imagine Reason (talk) 07:06, 28 March 2008 (UTC)

You know Bézout's lemma, right? Algebraist 15:23, 28 March 2008 (UTC)

Three calculus problems

Hi, I'm studying for a test tomorrow and there are three practice problems I haven't been able to do. I was wondering if anyone could show me how so that I will be able to solve analogous problems tomorrow.

1. ∫e^-x/(1 + e^-x)dx. I can't figure out what to make my u. Should I be looking for a du/u kind of pattern so that it will evaluate to ln |u| + c?

2. ∫₁³((e^3/x)/x²). My confusion is the same as in problem 1.

3. For this one I actually got an answer. I'm supposed to use logarithmic differentiation to evaluate dy/dx. The problem is y = x√(x² - 1). Here is my work:
ln y = ln(x√(x² - 1))
ln y = ln x + ln √(x² - 1)
ln y = ln x + (1/2)ln (x² - 1)
(1/y)(dy/dx) = 1/x + (1/2)(1/(x² + 1))·2x
(1/y)(dy/dx) = 1/x + (1/(x² + 1))·x
(1/y)(dy/dx) = 1/x + x/(x² + 1)
dy/dx = y(1/x + x/(x² + 1))
substituting from the original problem dy/dx = x√(x² - 1)(1/x + x/(x² + 1)) The back of the book says that dy/dx = (2x² - 1)/√(x² - 1). Is this form equivalent to what I got?

Any help is greatly appreciated. Thanks again, anon. —Preceding unsigned comment added by 141.155.126.183 (talk) 04:05, 27 March 2008 (UTC)

• Your intuition on #1 is correct. The key to a u-sub (at least at an introductory level) is usually to make the more complicated part of the integrand your u. #2 doesn't look like it involves du/u at all, actually... think about the exponent on the e and also the fact that the x^2 happens to be in the denominator, meaning it's actually x^-2. Play around with that a bit and see what happens. :) --Kinu ^t/_c 05:49, 27 March 2008 (UTC)
• Oh, and on #3... the process is correct, but when you took the derivative, the x^2-1 on the right became an x^2+1 for the remainder of the problem. Switching the sign to a negative as it should be yields an equivalent answer to the book's. Good work! --Kinu ^t/_c 05:56, 27 March 2008 (UTC)
  • For #1, it can involve substitution by letting u equal the denominator. Then one gets $\displaystyle \int \displaystyle {\frac {u-1}{u}}\displaystyle {\frac {1}{1-u}}\,du$ which can obviously be simplified ($a-b=-(b-a)$). x42bn6 Talk Mess 18:03, 27 March 2008 (UTC)

Why not just let u be $e^{-x}+1$ so $du=-e^{-x}dx$? A math-wiki (talk) 21:10, 27 March 2008 (UTC)

Isn't that what x42bn6 said, just written out with an extra intermediate step? --Tango (talk) 23:43, 27 March 2008 (UTC)

Though he says "it can involve substitution by letting u equal the denominator." His integrand suggests a different choice for u, if he had substituted directly, he should have gotten $-\int {\frac {du}{u}}$ right off the bat. A math-wiki (talk) 03:10, 28 March 2008 (UTC)

No, that is the same u. The ${\frac {u-1}{u}}$ part is from the ${\frac {e^{-x}}{e^{-x}+1}}$ part and the ${\frac {1}{1-u}}\,du$ part is from the dx. You then cancel to get what you got. It's the same method, you've just skipped a step in the substitution (I would have done, too) by observing that you can just group $e^{-x}\,dx$ together and substitute it in one go. --Tango (talk) 15:58, 28 March 2008 (UTC)

I suppose, though it seems unnecessarily complicated to do it that way. It has been 2 years since I did much Calculus. A math-wiki (talk) 21:39, 28 March 2008 (UTC)

Yeah, I jumped a "step", if you will - I usually omitted it because it saves typing. I don't know what "du/u" meant in any case - I was never a fan of treating "dx" as a variable. x42bn6 Talk Mess 20:43, 29 March 2008 (UTC)

Well, the fact is that Leibniz notation works, and you can often save a lot of effort by abusing it. Even without doing so, $\int {\frac {du}{u}}$ is just another way to write $\int {\frac {1}{u}}du$. -- Meni Rosenfeld (talk) 21:08, 29 March 2008 (UTC)

I thought it was akin to ${\frac {dy}{dx}}=x\Rightarrow dy=xdx$ with the dx and dy treated as variables, not that notation. x42bn6 Talk Mess 22:04, 29 March 2008 (UTC)

Factorials and Primes

I've noticed this:

• Prime factors of 3!-2!-1!-0!: 2
• Prime factors of 4!-3!-2!-1!-0!: 2, 7
• Prime factors of 5!-4!-3!-2!-1!-0!: 2, 43
• Prime factors of 6!-5!-4!-3!-2!-1!-0!: 2, 283
• Prime factors of 7!-6!-5!-4!-3!-2!-1!-0!: 2, 2083
• Prime factors of 8!-7!-6!-5!-4!-3!-2!-1!-0!: 2, 17203 (mmm, these all end in 3)
• Prime factors of 9!-...: 2, 11, 37, 389 (this and the three below don't work)
• Prime factors of 10!-...: 2, 1609843
• Prime factors of 11!-...: 2, 853, 21031
• Prime factors of 12!-...: 2, 23, 9457531
• Prime factors of 13!-...: 2, 101, 1523, 18541
• Prime factors of 14!-...: 2, 40214157043 (bingo, it's working again)
• Prime factors of 15!-...: 2, 606873049843
• Prime factors of 16!-...: 2, 9760593625843
• Prime factors of 17!-...: 2, 829, 201063350767 (stopped again)
• Prime factors of 18!-...: 2, 67, 397, 168143, 673499
• Prime factors of 19!-...: 2, 57432357441241843 (works again)
• Prime factors of 20!-...: 2, 1152238261120729843
• so on...

Do you guys have any explanations for this, or why the ones with only two prime factors have 2 and then some number ending in 3?

--wj32 ^t/c 07:05, 27 March 2008 (UTC)

Also, it doesn't seem to work from 21!-... onwards... --wj32 ^t/c 07:10, 27 March 2008 (UTC)

Put f(n) = n! − (n−1)! − (n-2)! ... − 0!. For n ≥ 5, n! ≡ 0 mod 20. That means that f(n) ≡ -(4! + 3! + 2! + 1! + 0!) ≡ 6 mod 20, so f(n)/2 ≡ 3 mod 10. So after factoring out 2, you have a number ending in 3. This also works for f(21) = 2·24264807338798809843 (where the larger factor is a multiple of 479) and onwards.  --Lambiam 08:01, 27 March 2008 (UTC)

f(n)/2 never has a prime factor ≤ n. It is a proved prime for n = 4, 5, 6, 7, 8, 10, 14, 15, 16, 19, 20, 32, 90, 353, 544 (1254 digits), and a gigantic probable prime for n = 3479 (10812 digits), 3602 (11249 digits), 3800 (11955 digits), and no other n below 4000. Guess the reason for my username! PrimeHunter (talk) 15:56, 27 March 2008 (UTC)

Thanks, that cleared everything up for me! --wj32 ^t/c 05:53, 28 March 2008 (UTC)

Finding Non-real solutions of systems of equations

Do any of you know of any methods for finding solutions of the form (a+bi,c+di) for systems of equations? Thanks in advance. A math-wiki (talk) 09:49, 27 March 2008 (UTC)

How about exactly like finding real solutions, only without assuming the solutions are real?

More specifically, methods for solving systems of linear equations like Gaussian elimination work over any field, complexes included. So you really just do everything normally, but if you have an underspecified system, you let the free variables range over the complexes rather than the reals.

If the equations are nonlinear, again you can use any of the known methods, such as extracting one unknown from an equation and substituting it in another. You only have to remember that when inverting a function, you need to keep in mind nonreal preimages. So, for $z^{2}=-1$ you don't conclude there is no solution, and for $e^{z}=2$ the solution is not just $\ln 2$ but $\ln 2+2\pi ki$.

In some cases you can simplify the problem by representing the complex unknown with real numbers, such as $z=a+bi$ or $z=r\mathrm {cis} \theta$, eliminating all instances of i, and proceeding with the mundane real equations.

If none of this helps with what you had in mind, you will have to specify it. -- Meni Rosenfeld (talk) 12:58, 27 March 2008 (UTC)

Well the system I'm particularly interested in is

${\begin{cases}{\frac {B^{2}}{4}}+2n+b=C\\{\frac {2Bn+Bb}{2}}=D\end{cases}}$

We know B,C, and D and we are trying to find b and n. This system comes from the latest addition to my userpage, solving Polynomials with Compositions of Functions. So long as this system has a solution, then the Quartic can be solved using the Composition of Function technique. A math-wiki (talk) 21:18, 27 March 2008 (UTC)

Hmm, when I solve for b in the first equation and substitute it into the second, n drops out... Any ideas?? A math-wiki (talk) 22:13, 27 March 2008 (UTC)

That's an underdetermined pair of linear equations, so Meni's remarks apply. There's an infinite family of solutions if 2D/B=C-B²/4, and no solutions otherwise. Just remember to allow free variables to range over C, as Meni said. Algebraist 22:40, 27 March 2008 (UTC)

Well, given that your equations are basically 2n + b = C - B^2 / 4 and 2n + b = 2D / B, that's not surprising. The tricky bit is working out what, if anything, went wrong. Confusing Manifestation(Say hi!) 22:42, 27 March 2008 (UTC)

That's simple: what went wrong is that Mathwiki's trying to express a general quartic as a composition of two quadratics. This is impossible. Algebraist 22:51, 27 March 2008 (UTC)

I thought we've covered that here (the anon OP there is A math-wiki, before creating an account). -- Meni Rosenfeld (talk) 23:27, 27 March 2008 (UTC)

That is where the system came from, but how to solve it over the complex plane was never discussed, since I didn't ask. A math-wiki (talk) 03:22, 28 March 2008 (UTC)

Weight of paper

Can anyone help me with an estimate of the weight, in kilograms, of a box (2500 sheets, or 5 reams) of paper of 80g/m2 (grammage)? Thanks if you can help. —Preceding unsigned comment added by 213.84.41.211 (talk) 14:19, 27 March 2008 (UTC)

We need to know the size of the paper, but given that it's easy. Assuming A4, which is 210mm × 297mm or 0.21m x 0.297m, so area per sheet = 0.06237 m²; multiply by 2500 sheets = 155.925 m²; at 80 gsm, this weighs 12474g, or 12.474kg, or about 27.5 pounds. (Plus the weight of the box of course.) AndrewWTaylor (talk) 14:38, 27 March 2008 (UTC)

Since A4 means the area of a sheet is (but for rounding) 2⁻⁴m², one A4 sheet of 80g/m² paper weighs 2⁻⁴m²·80g/m² = 5g. The difference caused by rounding, about 0.2%, is less than the difference between a dry and a humid day. Using this, you should be able to do the calculations in your head: 2500 × 5g = 2.5 × 5kg = 12.5kg.  --Lambiam 20:33, 27 March 2008 (UTC)

Ah, nice thinking, Lambiam. I see our article on Paper density makes the same observation about the weight of an 80 gsm A4 sheet (though without the qualification about rounding). AndrewWTaylor (talk) 08:31, 28 March 2008 (UTC)

Core 2 Exam Revision

Hello,

I am trying to do a question on my Core 2 exam revision paper (I'm doing A Level maths, first year), and I not sure how to go about it.

The question is

f(x) = x³-2x²+ax+b a and b are constants.

When f(x) is divided by (x-2), the remainder is 1. When f(x) is divided by (x+1), the remainder is 28.

Find the value of a and the value of b.

So I was wondering if anyone would offer a step by step guide on how to complete the question, I think it has something to do with the remainder theorem, but we haven't been taught how to deal wih two variables.

89.242.2.102 (talk) 14:28, 27 March 2008 (UTC)

You are told that "when f(x) is divided by (x-2), the remainder is 1". So you know that

$f(x)=(x-2)p(x)+1$

where p(x) is some polynomial in x. Setting x equal to 2 gives

$f(2)=(2-2)p(x)+1=1$

But we also know that

$f(2)=2^{3}-2(2^{2})+2a+b=2a+b$

so

$2a+b=1$

Carry out the same process with (x+1), setting x equal to -1 this time, and you will get a second equation in a and b. Solve the pair of simultaneous equations to find the values of a and b. Gandalf61 (talk) 14:50, 27 March 2008 (UTC)

Thank you for a super-fast response. Finally I can continue with my maths work, YAY!

89.242.2.102 (talk) 14:55, 27 March 2008 (UTC)

Actually just one more thought, when I put (x+1) in does the equation look like this:

f(-1) = (-1-2)p(x)+28=28  ?

89.242.2.102 (talk) 15:03, 27 March 2008 (UTC)

Not quite. Starting with "when f(x) is divided by (x+1), the remainder is 28", this means that

$f(x)=(x+1)q(x)+28$

where q(x) is some polynomial in x. Now set x equal to -1 ... Gandalf61 (talk)

So I get

f(-1) = (-1+1)q(x)+28 = 28

so what do I do next? I don't understand how you went from step 2:

$f(2)=(2-2)p(x)+1=1$

But we also know that

to step 3:

$f(2)=2^{3}-2(2^{2})+2a+b=2a+b$

so

$2a+b=1$

89.242.2.102 (talk) 16:03, 27 March 2008 (UTC)

We know that f(x) is defined as $f(x)=x^{3}-2x^{2}+ax+b$.

We don't know what a and b are yet, but we can just treat them as unknown constants. Substituting x=2 into this definition gives f(2)=2a+b.

But we also know, from using the remainder theorem, that f(2)=1. So this must mean that 2a+b=1.

If you want to follow this up further, I suggest you post a message on my talk page at User talk:Gandalf61, as other Ref Desk readers have probably lost interest in this topic ! Gandalf61 (talk) 16:15, 27 March 2008 (UTC)

Indeed kudos to you though for helping me out and all.

89.242.2.102 (talk) 16:26, 27 March 2008 (UTC)

Finding the solution

This is absolutely homework, and I can need a wee bit of guidance. If anyone can help by pointing me in the right direction, I would be very delighted.

I need to take the following: 1-1/4+1/27-1/256+1/3125-1/46656+1/823543 - and find a solution for it.

The solution is as follows:

7

(sumsign here) (-1)^(k+1) * (1/k^k)

k=1

How do I go from mere numbers to an algebraic expression? Thank you for your help. 81.93.102.185 (talk) 16:47, 27 March 2008 (UTC)

Since it's just a sum of 7 rational numbers, just get a common denominator and add them together, no? –King Bee ^{(τ • γ)} 18:55, 27 March 2008 (UTC)

My assignment is asking me to find the algebraic solution that is (7, k=1) (-1)^(k+1) * (1/k^k). Herein lies my problem, since I do not know of any procedures to find the algebraic expression. 81.93.102.185 (talk) 19:12, 27 March 2008 (UTC)

Note that 2^2 = 4, 3^3 = 27, 4^4 = 256, 5^5 = 3125, etc. SmaleDuffin (talk) 19:56, 27 March 2008 (UTC)

I don't understand your problem. You say you need to find the solution, which is as follows:

$\sum _{k=1}^{7}{\frac {{(-1)}^{k+1}}{k^{k}}}.$

If that is the solution, then that is the solution, so you have found it. Why do you need to look further? Is the assignment you have in written form? What exactly does it say?  --Lambiam 08:58, 28 March 2008 (UTC)

I think that the OP has found this expression in an answers sheet and asks how he could have found it on his own. -- Meni Rosenfeld (talk) 10:02, 28 March 2008 (UTC)

quick sum question

What is 6,916,214 + 1,254,934,496 + 48,322,260 + 851? Thanks.

A somewhat meaningless number, which is currently in the range of 200,000,000 - 250,000,000. What's 5 apples + 27 Tuesdays? Confusing Manifestation(Say hi!) 22:26, 27 March 2008 (UTC)

Edits over time however, would give us the power of Wikipedia ^_^ Mac Davis (talk) 23:50, 27 March 2008 (UTC)

This is the sum: 220,878,069...Oh wow. This'll change quickly. Spencer^T♦C 00:09, 28 March 2008 (UTC)

1310173821. I win! ;) — Kieff | Talk 00:50, 28 March 2008 (UTC)

I got the same thing as Kieff. Tesseran (talk) 01:37, 28 March 2008 (UTC)

My calculator says Kieff's on the money ;) A math-wiki (talk) 03:26, 28 March 2008 (UTC)

Algebra Question

This is the original problem: -2z^2+32. My teacher says the answer is -2(x-4)(x+4). How does she come up with this? —Preceding unsigned comment added by 64.119.61.7 (talk) 22:58, 27 March 2008 (UTC)

If you factor out -2, you get -2(z²-16) because of the distributive property. z²-16 is the difference of perfect squares. You end up with -2(x-4)(x+4). Paragon12321 (talk) 23:43, 27 March 2008 (UTC)

Presumably all of the variables are meant to be z's, or else x's, not both. —Bkell (talk) 01:49, 28 March 2008 (UTC)